Question

QUESTION 1 0.5 points Save Answer An energy diagram for the formation of crystalline potassium fluoride is shown below: A kig) +E+F(g) Kig)+F) Kig)+F) K(g) +1/25212 K()+1/2FTI KF) and the electron affinity of fluorine is labeled by the number In this diagram, the sublimation energy of potassium is labeled by the number {Enter as a whole number.) QUESTION 2 0.5 points Save Answer What volume (in Llis accupied by 2.39 moles of gas at -22.4 °C with a pressure of 948 mm Hg? QUESTION 3 0.5 points Save Answer Alesk is filed with 1.34 moles of a gas at 18.5 K and 438.6 mm Hg. The flask is then opened and an additional 1.39 moles are added. The temperature of the flask is then changed to 100.6 K. What is the pressure fin atm of the flask under these final conditions?

Answer 1

2.1 In the from given Solid diagram to gas Potassium change it's stale in labeled o and Sublimation Potassium election in labeled ③ loose the L election affinity 8.2 mole, (n) = 2.39 Pressure (P1 = gudmmung = gud atm Temp. (T) = -22.4°C = (-22.4 +273115), k = 250.75 k Cras constant (R) = 0.082 atm. L . atm = 760 mmHg Ideal gas hac pr=nRT = v= nRT 760 Molik 390.082 x 250.75 - 49.141 = 39.40 L Р V= 2. 39 0.08 x. (948/760) 1:247

DATE 1:2 0:3 Intial mole (n) = 1.34 Temp (T) = 18.5k - Pression (P) = 438.6 mmHg = 438.6 colm 760 PV = nRT V= HRY = 1.348010 82 x 1815 e (4384760) V = 2.0327 = 3.522 , L 0.5771 Now we add 1.39 mole in same Volume Container than total Volume (V) = 3:522,L total mole (n) = 1.39+ 1.34 = 2.73 - Temp (T1= 100.6k than P = RT = 2.7380 1082 x 100.6 v 3.522 P = 6.3941, atm = 6.3941 8760 mm Hg - - 4859.516 mm Hg . I atm = 760 mm Hg